exercise 1.108 (Greece 2008) [BOMC]

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exercise 1.108 (Greece 2008) [BOMC]

Unread post by nafistiham » Sun Oct 30, 2011 9:28 pm

can we use anything else than "the helpful inequality"

for $x_{1},x_{2},x_{3},\cdot \cdot \cdot x_{n}$ positive integers, prove that

\[(\frac{x_{1}^{2}+x_{2}^{2}+\cdot \cdot \cdot +x_{n}^{2}}{x_{1}+x_{2}+\cdot \cdot \cdot +x_{n}})^{\frac{kn}{t}}\geq x_{1}\cdot x_{2}\cdot \cdot \cdot \cdot \cdot x_{n}\],

where $k=$ max {${{x_{1}, x_{2},\cdot \cdot \cdot ,x_{n}}}$} and $t=$ min {${{x_{1}, x_{2}, \cdot \cdot \cdot, x_{n}}}$}.
under which conditions the equality holds?
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: exercise 1.108 (Greece 2008) [BOMC]

Unread post by *Mahi* » Mon Oct 31, 2011 10:38 am

In fact you can use Cauchy or QM-AM inequalities to simplify this problem.
\[ ( \sum_{i=1} ^n x_i^2 )(\sum_{i=1} ^n 1) \geq (\sum_{i=1}^n x_i)^2 \]
Or, \[\frac { \sum_{i=1} ^n x_i^2} {\sum_{i=1}^n x_i} \geq \frac {\sum_{i=1}^n x_i}{n}\]
\[\sqrt {\frac { \sum_{i=1} ^n x_i^2} n} \geq \frac {\sum_{i=1}^n x_i}{n}\]
Or, \[\frac { \sum_{i=1} ^n x_i^2} {\sum_{i=1}^n x_i} \geq \frac {\sum_{i=1}^n x_i}{n}\]
Then solve the last part as you please.
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Re: exercise 1.108 (Greece 2008) [BOMC]

Unread post by Hasib » Mon Oct 31, 2011 3:15 pm

But, as far as i proved,

\[(\frac{\sum_{i=1}^{n} x_i^2}{\sum_{i=1}^{n} x_i})^n \geq x_1\cdot \cdot x_n\]

but, how raise the power $\frac{k}{t}$?
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